Question

The velocity of a particle traveling along a straight line is v = Vo − ks where k is constant. If s=0 when t=0, determine the position and acceleration of the particle as a function of time.

Answer 1

Answer:

acceleration of the particle as a function of time is given as

$a(t)=-\frac{ks}{t}$

and position of the particle as a function of time is given as

$x(t)=ut+\frac{at^2}{2} \\x(t)=V_{0}t-\frac{\frac{ks(t)}{t}t^2}{2}$

$x(t)=V_{0}t-\frac{kst}{2}$

Explanation:

we have given,

$v=v_{o}− ks$

at t=0, s=0

therefore initial velocity,u

u=v=$V_{o}$

v=u+at.........(1)

we know from first equation of motion

let acceleration at time t be a(t), and position be x(t)

put value of v,u and a in equation 1

we got,

$V_{o}-ks=V_{o}+a(t)t$

Therefore,acceleration of the particle as a function of time is given as

$a(t)=-\frac{ks}{t}$

and position of the particle as a function of time is given as

$x(t)=ut+\frac{at^2}{2} \\x(t)=V_{0}t-\frac{\frac{ks}{t}t^2}{2}$

$x(t)=V_{0}t-\frac{kst}{2}$